哪里（设置）IN（设置）

Question

编辑：我似乎不正确地问这个问题。

我试图找到一种方法来查询，如果一套在另一套可用。例如：

SELECT * FROM something 
WHERE (1, 3) IN (1, 2, 3, 4, 5) 

在这种情况下，1 & 3是在集合（1，2，3，4，5）。又如：

SELECT * FROM something 
WHERE (1, 3) IN (1, 5, 7, 9); 

在这种情况下，1 & 3在设定不（1，5，7，9），所以没有什么应该从表中可以拉动。

Answer 1

新的要求（根据sqlfiddle.com/#!9/f36d92/2）：

 
# The goal is to write a query that will select all exercises 
# that the user has the correct equipment for, where the pre-defined 
# set is the id's of the equipment the user has. 

# For example, let's assume the user has equipment (1, 4) 
# The exercise "Curls" should be pulled from the table, as the user has all 
# of the required equipment based on the exercise_requirements table. 
# while "Wrecking Ball" is not returned as the user only has a portion of the 
# required equipment. 

# If the user's equipment was (1, 3, 4) then both "Curls" and "Wrecking ball" 
# would be returned from the exercises table, as the user has the required equipment 
# for both exercises. 

#---- 

#Below is my take on your query. 

SELECT ex.* FROM exercises ex 
WHERE ex.id IN (
    SELECT exercise_id FROM exercise_requirements 
    WHERE ex.id IN (1, 4) 
    GROUP BY exercise_id 
    HAVING COUNT(distinct exercise_id) = 3 
); 

SOLUTION：

您在这里混淆了一些标识。这将更接近：

SELECT ex.* FROM exercises ex 
WHERE ex.id IN (
    SELECT exercise_id FROM exercise_requirements 
    WHERE equipment_id IN (1, 4) 
    GROUP BY exercise_id 
    HAVING COUNT(distinct equipment_id) = 2 
); 

但仍然这个查询是反之亦然。我们不想知道用户的所有设备是否都存在于一套锻炼所需的设备中，但是否需要进行锻炼的整套设备是在用户的设备中找到的。

可能最简单的写法是：合计exercise_requirements每exercise_id并检查没有equipment_id是用户不需要的。

select * 
from exercises 
where id in 
(
    select exercise_id 
    from exercise_requirements 
    group by exercise_id 
    having sum(equipment_id not in (1, 4)) = 0 
); 

您更新小提琴：http://sqlfiddle.com/#!9/f36d92/5

Answer 2

您可以使用此

SELECT u.* 
FROM users u 
INNER JOIN completed_levels cl 
ON cl.user_id = u.id 
WHERE cl.id IN (1, 5, 7); 

或者使用EXISTS作为link从@DanFromGermany

Answer 3

您可以使用Case做出Sum将与1增加1内的每个级别，5 & 7。

SELECT A.* FROM users AS 
INNER JOIN 
(
SELECT U.id, 
SUM(CASE WHEN 
     (
     A.completed_levels = 1 
     OR A.completed_levels = 5 
     OR A.completed_levels = 7 
    ) THEN 1 ELSE 0 END 
    ) AS RN 
FROM completed_levels A 
INNER JOIN users U ON A.user_id = U.id 
GROUP BY U.id 
) B ON A.id = B.id 
WHERE B.RN = 3 -- Those users have completed level 1, 5 & 7 will have RN = 3 only 

Answer 4

您正在使用IN子句相关的子查询（即，子查询引用u.id）。这不是我们如何使用它。 IN子句适用于不相关的子查询;如果您需要相关子查询，请改用EXISTS。对于您的问题，非相关子查询就足够了，所以用IN相应：

select * 
from users 
where u.id in (select user_id from completed_levels where id in (1, 5, 7); 

如果用户必须拥有所有级别：

select * 
from users 
where u.id in (select user_id from completed_levels where id = 1 
    and u.id in (select user_id from completed_levels where id = 5 
    and u.id in (select user_id from completed_levels where id = 7; 

这样的问题通常是更好地与聚集，从而解决不必一次又一次查询同一表：

select * 
from users 
where u.id in 
(
    select user_id 
    from completed_levels where id in (1, 5, 7) 
    group by user_id 
    having count(distinct id) = 3 
); 

Answer 5

注意：这回答了原来的问题，这似乎与OP修改后的问题无关。

你可以得到谁完成所有三个级别使用的用户：

SELECT cl.user_id 
FROM completed_levels cl 
WHERE cl.id IN (3, 5, 7) 
GROUP BY cl.user_id 
HAVING COUNT(DISTINCT cl.id) = 3; 

（注：DISTINCT是没有必要的，如果给定用户的ID是唯一的）

然后，您可以得到你想要使用JOIN或类似的结构是什么：

SELECT u.* 
FROM users u JOIN 
    (SELECT cl.user_id 
     FROM completed_levels cl 
     WHERE cl.id IN (3, 5, 7) 
     GROUP BY cl.user_id 
     HAVING COUNT(DISTINCT cl.id) = 3 
    ) cu 
    ON cl.user_id = u.id;